1. The radioisotope Cobalt-60 is used in cancer therapy. The half-life of this isotope is 5.27 years.
Which equation determines the percent of an initial amount of the isotope remaining after t years?
2. What are the real and imaginary parts of the complex number?
The real part:
The imaginary part:
My Answer- Real part=9, Imag. part=7i
What is the simplest radical form of the expression?
Thanks so much
The correct equation in the first part should be
A = 100(1/2)t/5.27
To see that this is correct....let t = 5.27 and we have that
A = 100(1/2) 5.27/5.27 = 100 (1/2)1 = 100 (1/2) = 50% left after 5.27 years, the half-life
Your second answer is correct
I can't tell exactly what the third one is....sorry....!!!